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VEDIC MATHEMATICS : Arithmetic Operations

VEDIC MATHEMATICS : Arithmetic Operations. T. K. Prasad http://www.cs.wright.edu/~tkprasad. Positional Number System. 4 3 2 1 0 = 4 * 10,000 + 3 * 1,000 + 2 * 100 + 1 * 10 + 0. Two Digit Multiplication (above the base) using Vedic Approach . Method : Vertically and Crosswise Sutra

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VEDIC MATHEMATICS : Arithmetic Operations

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  1. VEDIC MATHEMATICS : Arithmetic Operations T. K. Prasad http://www.cs.wright.edu/~tkprasad Arithmetic Operations Revisited

  2. Positional Number System 43210 = 4 * 10,000 + 3 * 1,000 + 2 * 100 + 1 * 10 + 0 Arithmetic Operations Revisited

  3. Two Digit Multiplication (above the base) using Vedic Approach Method : Vertically and Crosswise Sutra Correctness and Applicability Arithmetic Operations Revisited

  4. Method: Multiply 13 * 12 • Write the first number to be multiplied and excess over 10 in the first row, and the second number to be multiplied and excess over 10 in the second row. 133 122 Arithmetic Operations Revisited

  5. 13 3 12 2 • To determine the 3-digit product: • addcrosswise to obtain the left digits • (13 + 2) = (12 + 3) = 15 • and • multiply the excess vertically to obtain the right digit. • (3 * 2) = 6 • 13 * 12 = 156 Arithmetic Operations Revisited

  6. Another Example • 12 * 14 = • 12 2 • 14 4 • 16 8 • 12 * 14 = 168 Arithmetic Operations Revisited

  7. Questions • Why do both crosswise additions yield the same result? • Why does this method yield the correct answer for this example? • Does this method always work for any pair of numbers? Arithmetic Operations Revisited

  8. Proof Sketch • (12 + 4) = (14 + 2) = 16 • Why are they same? • That is, the sum of first number and excess over 10 of the second number, and …. • (12 + (14 – 10)) = (12+14 – 10) = (26 – 10) = 16 • (14 + (12 – 10)) = (14+12 – 10) = (26 – 10) = 16 Arithmetic Operations Revisited

  9. 12 = (10 + 2) 14 = (10 + 4) 12 * 14 = (10 + 2) * 14 = 10 * 14 + 2 * 14 = 10 * 14 + 2 * (10 + 4) = 10 * 14 + 2 * 10 + (2 * 4) = 10 * (14+2) + 8 = 10 * 16 + 8 = 168 12 = (10 + 2) 14 = (10 + 4) 12 * 14 = 12 * (10 + 4) = 12 * 10 + 12 * 4 = 12 * 10 + (10 + 2) * 4 = 12 * 10 + 10 * 4+ (2 * 4) = 10 * (12+ 4) + 8 = 10 * 16 + 8 = 168 Correctness Argument:Two possibilities Right digit [Vertical Product] Right digit [Vertical Product] Left digits [Crosswise Addition] Left digits [Crosswise Addition] Arithmetic Operations Revisited

  10. Another Example • 15 * 12 15 5 12 2 17 10 18 0 Arithmetic Operations Revisited

  11. Yet Another Example • 17 * 15 17 7 15 5 22 35 22+35 25 5 Need proof to feel comfortable! Arithmetic Operations Revisited

  12. Method: Multiply 113 * 106 • Write the first number to be multiplied and excess over 100 in the first row, and the second number to be multiplied and excess over 100 in the second row. 11313 1066 Arithmetic Operations Revisited

  13. 113 13 106 6 • To determine the 5-digit product: • addcrosswise to obtain the left digits • (113 + 6) = (106 + 13) = 119 • and • multiply the excess vertically to obtain the right digits. • (13 * 6) = 78 • 113 * 106 = 11978 Arithmetic Operations Revisited

  14. Questions • Why do both crosswise additions yield the same result? • Why does this method yield the correct answer for this example? • Does this method always work for any pair of 3 digit numbers? Arithmetic Operations Revisited

  15. Proof Sketch • (113 + 6) = (106 + 13) = 119 • Why are they same? • (113 + (106 – 100)) = (113 + 106 – 100) = 119 • (106 + (113 – 100)) = (106 + 113 – 100) = 119 Arithmetic Operations Revisited

  16. 113 = (100 + 13) 106 = (100 + 6) 113 * 106 = 113 * (100 + 6) = 113 * 100 + (100 + 13) * 6 = 113 * 100 + 100 * 6+ (13 * 6) = 100 * (113+ 6) + 78 = 100 * 119 + 78 = 11978 113 = (100 + 13) 106 = (100 + 6) 113 * 106 = (100 + 13) * 106 = 100 * 106 + 13 * (100 + 6) = 100 * 106 + 13 * 100 + (13 * 6) = 100 * (106+ 13) + 78 = 100 * 119 + 78 = 11978 Correctness of Product :Two possibilities Right digits [Vertical Product] Right digits [Vertical Product] Left digits [Crosswise Addition] Left digits [Crosswise Addition] Arithmetic Operations Revisited

  17. Another Example Breakdown?! • 160 * 180 160 60 180 80 240 4800 288 00 • Note that, the product of the excess over 100 has more than two digits. However, the weight associated with 240 and 48 are both 100, and hence they can be combined. Arithmetic Operations Revisited

  18. Yet Another Example • 190 * 199 190 90 199 99 289 8910 289+89 10 378 10 This approach is validwith suggested modifications! Breakdown?! Arithmetic Operations Revisited

  19. More Shortcuts Arithmetic Operations Revisited

  20. Quick squaring of numbers that end in 5 Proof: Let the two digit number be written as D5. D5 * D5 = (D*10 + 5) * (D*10 + 5) = (D*D*100) + (D*2*50) + 5*5 = (D*(D+1))*100 + 25 • 15 * 15 = 225 = (1*2) (5*5) • 75 * 75 = 5625 = (7*8) (5*5) • 95 * 95 = 9025 = (9*10) (5*5) Arithmetic Operations Revisited

  21. Quick Multiplication : Special Case Proof: Let two digit numbers be AB and AC. AB * AC = (A*10 + B) * (A*10 + C) = (A*A*100) + (A*10*(B+C)) + B*C = (A*A)*100 + (A)*(B+C)*10 + (B*C) For B+C=10, this reduces to A*(A+1)*100 + B*C For A=12, B=8 and C=2, this reduces to (12)*(13)*100 + 16 = 15616 Arithmetic Operations Revisited

  22. Quicking squaring of numbers that begin with 5 Proof: Let the two digit number be written as 5D. 5D * 5D = (50 + D) * (50 + D) = (25 + D)*100 + (D*D) • 51 * 51 = (5*5+1)*100 + (1*1) = 2601 • 57 * 57 = (5*5+7) *100 + (7*7) = 3249 • 59 * 59 = (5*5+9) *100 + (9*9) =3481 Arithmetic Operations Revisited

  23. Quick squaring of two digit numbers Proof: Let two digit numbers be AB. AB * AB = (A*10 + B) * (A*10 + B) = (A*A)*100 + 2*(A*10)*B + B*B = (A*A)*100 + 20*(A*B) + (B*B) For AB=79, this reduces to 4900+20*63+81 = 4981+1260 =6241 For AB=116, this reduces to 12100+20*66+36 = 12136+1320 =13456 Arithmetic Operations Revisited

  24. Generalized Multplication Using Working Base Arithmetic Operations Revisited

  25. 23 +3 24 +4 • To determine the product, choose working base as 20: • addcrosswise to obtain the left digits with weight 20 • (23 + 4) = (24 + 3) = 27 • multiply the excess vertically to obtain the right digits. • (3 * 4) = 12 • 23 * 24 = 27 * 20 + 12 • = 540 + 12 23 * 24 = 552 Arithmetic Operations Revisited

  26. 723 +23 724 +24 • To determine the product, choose working base as 700: • addcrosswise to obtain the left digits with weight 700 • (723 + 24) = (724 + 23) = 747 • multiply the excess vertically to obtain the right digits. • (23 * 24) = 552 • 723 * 724 = 747 * 700 + 552 • = 522900 + 552 723 * 724 = 523452 Arithmetic Operations Revisited

  27. 783 -17 775 -25 • To determine the product, choose working base as 800: • addcrosswise to obtain the left digits with weight 800 • (783 - 25) = (775 - 17) = 758 • multiply the excess vertically to obtain the right digits. • (17 * 25) = 425 • 783 * 775 = 758 * 800 + 425 • = 606400 + 425 783 * 775 = 606825 Arithmetic Operations Revisited

  28. 532 +32 472 -28 • To determine the product, choose working base as 1000/2: • addcrosswise to obtain the left digits with wt. 1000/2 • (532 - 28) = (472 + 32) = 504 • multiply the excess vertically to obtain the right digits. • (+32) * (-28) = 896 • 532 * 472= (504 / 2)*1000 + (104 -1000) • = 252000 + 104 - 1000 532 * 472= 251104 Arithmetic Operations Revisited

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